Why the alternate form of $frac{sqrt3}2+frac i 2$ is $sqrt[6]{-1}$?
$begingroup$
The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)
What are the arithmetic actions that gets us from the former to the latter?
algebra-precalculus complex-numbers arithmetic
edited Dec 13 '18 at 22:51
gimusi
1
asked Dec 13 '18 at 22:33
HeyJudeHeyJude
1636
$endgroup$
add a comment |
$begingroup$
The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)
What are the arithmetic actions that gets us from the former to the latter?
algebra-precalculus complex-numbers arithmetic
edited Dec 13 '18 at 22:51
gimusi
1
asked Dec 13 '18 at 22:33
HeyJudeHeyJude
1636
$endgroup$
2
$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35
$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44
$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46
add a comment |
$begingroup$
The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)
What are the arithmetic actions that gets us from the former to the latter?
algebra-precalculus complex-numbers arithmetic
edited Dec 13 '18 at 22:51
gimusi
1
asked Dec 13 '18 at 22:33
HeyJudeHeyJude
1636
$endgroup$
The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)
What are the arithmetic actions that gets us from the former to the latter?
algebra-precalculus complex-numbers arithmetic
algebra-precalculus complex-numbers arithmetic
edited Dec 13 '18 at 22:51
gimusi
1
asked Dec 13 '18 at 22:33
HeyJudeHeyJude
1636
edited Dec 13 '18 at 22:51
gimusi
1
asked Dec 13 '18 at 22:33
HeyJudeHeyJude
1636
edited Dec 13 '18 at 22:51
gimusi
1
edited Dec 13 '18 at 22:51
gimusi
1
edited Dec 13 '18 at 22:51
gimusi
1
1
asked Dec 13 '18 at 22:33
HeyJudeHeyJude
1636
asked Dec 13 '18 at 22:33
HeyJudeHeyJude
1636
asked Dec 13 '18 at 22:33
HeyJudeHeyJude
1636
1636
2
$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35
$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44
$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46
add a comment |
2
$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35
$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44
$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46
2
2
$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35
$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35
$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44
$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44
$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46
$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.
The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.
The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
$$
omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
$$
Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.
edited Dec 14 '18 at 12:06
Scientifica
6,37641335
answered Dec 13 '18 at 22:41
Martin ArgeramiMartin Argerami
124k1177176
$endgroup$
add a comment |
$begingroup$
$$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$
The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.
answered Dec 13 '18 at 22:39
Mike EarnestMike Earnest
20.9k11951
$endgroup$
add a comment |
$begingroup$
We simply have that
$$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$
and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.
Refer also to:
- Principal $n$th root of a complex number
answered Dec 13 '18 at 22:37
gimusigimusi
1
$endgroup$
1
$begingroup$
This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
$endgroup$
– Blue
Dec 13 '18 at 22:44
$begingroup$
That's the principal root! Ok I add something more on that.
$endgroup$
– gimusi
Dec 13 '18 at 22:46
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038663%2fwhy-the-alternate-form-of-frac-sqrt32-frac-i-2-is-sqrt6-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.
The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.
The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
$$
omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
$$
Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.
edited Dec 14 '18 at 12:06
Scientifica
6,37641335
answered Dec 13 '18 at 22:41
Martin ArgeramiMartin Argerami
124k1177176
$endgroup$
add a comment |
$begingroup$
That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.
The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.
The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
$$
omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
$$
Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.
edited Dec 14 '18 at 12:06
Scientifica
6,37641335
answered Dec 13 '18 at 22:41
Martin ArgeramiMartin Argerami
124k1177176
$endgroup$
add a comment |
$begingroup$
That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.
The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.
The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
$$
omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
$$
Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.
edited Dec 14 '18 at 12:06
Scientifica
6,37641335
answered Dec 13 '18 at 22:41
Martin ArgeramiMartin Argerami
124k1177176
$endgroup$
That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.
The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.
The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
$$
omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
$$
Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.
edited Dec 14 '18 at 12:06
Scientifica
6,37641335
answered Dec 13 '18 at 22:41
Martin ArgeramiMartin Argerami
124k1177176
edited Dec 14 '18 at 12:06
Scientifica
6,37641335
edited Dec 14 '18 at 12:06
Scientifica
6,37641335
edited Dec 14 '18 at 12:06
Scientifica
6,37641335
6,37641335
answered Dec 13 '18 at 22:41
Martin ArgeramiMartin Argerami
124k1177176
answered Dec 13 '18 at 22:41
Martin ArgeramiMartin Argerami
124k1177176
answered Dec 13 '18 at 22:41
Martin ArgeramiMartin Argerami
124k1177176
124k1177176
add a comment |
add a comment |
$begingroup$
$$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$
The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.
answered Dec 13 '18 at 22:39
Mike EarnestMike Earnest
20.9k11951
$endgroup$
add a comment |
$begingroup$
$$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$
The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.
answered Dec 13 '18 at 22:39
Mike EarnestMike Earnest
20.9k11951
$endgroup$
add a comment |
$begingroup$
$$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$
The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.
answered Dec 13 '18 at 22:39
Mike EarnestMike Earnest
20.9k11951
$endgroup$
$$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$
The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.
answered Dec 13 '18 at 22:39
Mike EarnestMike Earnest
20.9k11951
answered Dec 13 '18 at 22:39
Mike EarnestMike Earnest
20.9k11951
answered Dec 13 '18 at 22:39
Mike EarnestMike Earnest
20.9k11951
answered Dec 13 '18 at 22:39
Mike EarnestMike Earnest
20.9k11951
20.9k11951
add a comment |
add a comment |
$begingroup$
We simply have that
$$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$
and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.
Refer also to:
- Principal $n$th root of a complex number
answered Dec 13 '18 at 22:37
gimusigimusi
1
$endgroup$
1
$begingroup$
This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
$endgroup$
– Blue
Dec 13 '18 at 22:44
$begingroup$
That's the principal root! Ok I add something more on that.
$endgroup$
– gimusi
Dec 13 '18 at 22:46
add a comment |
$begingroup$
We simply have that
$$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$
and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.
Refer also to:
- Principal $n$th root of a complex number
answered Dec 13 '18 at 22:37
gimusigimusi
1
$endgroup$
1
$begingroup$
This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
$endgroup$
– Blue
Dec 13 '18 at 22:44
$begingroup$
That's the principal root! Ok I add something more on that.
$endgroup$
– gimusi
Dec 13 '18 at 22:46
add a comment |
$begingroup$
We simply have that
$$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$
and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.
Refer also to:
- Principal $n$th root of a complex number
answered Dec 13 '18 at 22:37
gimusigimusi
1
$endgroup$
We simply have that
$$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$
and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.
Refer also to:
- Principal $n$th root of a complex number
answered Dec 13 '18 at 22:37
gimusigimusi
1
edited Dec 13 '18 at 22:49
answered Dec 13 '18 at 22:37
gimusigimusi
1
answered Dec 13 '18 at 22:37
gimusigimusi
1
answered Dec 13 '18 at 22:37
gimusigimusi
1
1
1
$begingroup$
This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
$endgroup$
– Blue
Dec 13 '18 at 22:44
$begingroup$
That's the principal root! Ok I add something more on that.
$endgroup$
– gimusi
Dec 13 '18 at 22:46
add a comment |
1
$begingroup$
This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
$endgroup$
– Blue
Dec 13 '18 at 22:44
$begingroup$
That's the principal root! Ok I add something more on that.
$endgroup$
– gimusi
Dec 13 '18 at 22:46
1
1
$begingroup$
This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
$endgroup$
– Blue
Dec 13 '18 at 22:44
$begingroup$
This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
$endgroup$
– Blue
Dec 13 '18 at 22:44
$begingroup$
That's the principal root! Ok I add something more on that.
$endgroup$
– gimusi
Dec 13 '18 at 22:46
$begingroup$
That's the principal root! Ok I add something more on that.
$endgroup$
– gimusi
Dec 13 '18 at 22:46
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038663%2fwhy-the-alternate-form-of-frac-sqrt32-frac-i-2-is-sqrt6-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35
$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44
$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46